Dear All,Back to the Top
Sorry, but a need to make this basic query:
Could anyone say how to calculate the concentrations from a polinomial
regression (Y = ax2 + bx + c where: y = area ratio and x =
concentration (ng/mL) in a bioanalytical assay?
Thanks,
Daniel
P.S n I have tried to use Baskara theorema, but I could not get
adequate results.
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Daniel,
You can perform this calculation easily in an Excel spreadsheet after
you have obtained the polynomial expression that best describes the
relationship between the area ratio and the concentrations of your
calibrators by using the "Goal Seek" function in the drop down menu
under "Tools". You will be asked to input the following:
"Set cell:" You need to enter the designation of the cell in which the
polynomial equation is expressed
"To value:" You need to enter the numerical value of the area ratio
"By changing cell:" You need to enter the designation of the cell in
which you want the concentration to be reported.
After you have provided this information, click on "OK" in the dialog
box and the concentration will appear in the cell that you have
designated for the answer.
Sincerely,
Richard Sams
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Dear Daniel,
For a square polynomal function, you have a simple option in Excel:
1. Make a graph of your data (x-y type)
2. Add a trendline, use the option "put equation in graph"
For anything more complicated, you can use the solver add-in (needs
installation via add-in manager), which will do numerical regression for
basicly every equation. You just make a spreadsheet with your measured
data
Y. For e.g. an equation like Y = ax3 + bx2 + cx + d, you create fou
cells
where you type your guestimates for the parameters a-d. Based on these
parameters and your equation, you calculate the "theoretical" values of
Y(th) and the sum of square differences sum((Y-Y(th))^2). Then you
minimize
this sum with the solver with variation of the parameters (a-d).
good luck,
Michael
Michael Gassen
Head of Pharmacology and Preclinical Development
4SC AG Fon: +49 89 700763-0
Am Klopferspitz 19a Fax: +49 89 700763-29
82152 Martinsried E-Mail: gassen.-a-.4sc.com
Germany Internet: www.4sc.com
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Hi,
since C and Y are constant in your eqution,
why not try to resolve ax2 + bx + c -Y =0
You should end up on ax2 + bx + K =0 with K =c-Y since Y should be > to
c
you should end up on :ax2 + bx - K =0
Then resolve the equation for x with finding the two roots with one
might be
negative (that you reject) and the other one your concentration.
It would help to have the constants of your equation to know which
solution
is adequate.
Do not if this helps,
Regards,
Pascal
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Dear Daniel,
There will be 2 outcomes possible for such a quadratic function (so
using
'goal seek' may not be the best way to solve) - obviously, only one will
correspond to your concentration (you will know which is correct). To
work
out the two possible values for x, use the following formulae (where Y
= ax
2 + bx + c):
x1 = - (b/(2a)) - {square root [ ((Y-c)/a) + (b/(2a))2]}
x2 = - (b/(2a)) + {square root [ ((Y-c)/a) + (b/(2a))2]}
I hope this is helpful.
Cheers,
Dave
Dear members,Back to the Top
I am writing regarding a recent debateabout the use of polynomials for
calibration purpose. Let me add a few comments on it.
I haven't retained the original messages. However, if my recollection
is correct, several members provided the right answers. More
specifically, two solutions exist for ax2 + bx +c =0. By the way,
somebody wrote that one is positive and the other is negative. This is
not always right, both may be positive, or both negative.
The existence of 2 solutions illustrate an undesirable property of
polynomials in calibrations: they are non-monotonic functions. For this
reason, I would like to suggest 2 alternatives. Bothare based
onmonotonic functions.
1. One optionwould beto use linear regression on log-transformation
data (on both axes). LnY=a*LnX + b. This option, however,suffers 2
main (hidden) drawbacks. One is an underlying assumption of a
constant-CV error model, rarely applicable to bioanalytical data.
Another one is that the model is essentially forced through the origin.
This is obvious after antilog transformation: Y = exp(b)*X**a.The
main value of this approachis its straightforward implementation.
2. The second option would be tu use NONlinear regression on
UNtransformed data,usingthe monotonic model: Y = a*X**n + b. By
contrast with option 1, these is no specific assumption about the error
model (feel free to use the most appropriate weighting function, if
any) and the model is not forced through the origin thanks to the
intercept "b". On the other hand, implementation is not straightforward
asit requires a nonlinear regression module (e.g. Excel solver).
Enjoy the testing!
Henri
Henri Merdjan
Consultant to the Life Science Industries
PK and PK/PD specialist
34, rue des Bergers
75015 Paris
France
hmerdjan.-at-.wanadoo.fr
Back to the Top
Dear Henri,
Thank your for your suggestions for alternatives for the polynomial
regression.
Your comments with respect to the properties of linear regression
applied to log-transformed data are correct, but I do not fully agree
with your conclusions:
- the underlying assumption of a constant-CV error may be not fully
justified, but this also applies to the assumption of a constant error
underlying the use of linear regression on untransformed and unweighted
data. In real data, the situation is usually somewhere between.
- forcing the model through the origin is not a bad idea. In case one
knows that there a blank sample produces a response, one can make
corrections for it. Anyhow, if this is the case, also linear regression
may give rise to inaccuracies, and one has to be quite sure that the
calibration curve is linear. Please note that after log-transformation
the concentration zero is infinitely far away from the lowest
calibration sample. Nobody would ever make such an extrapolation! So it
is not really important that the calibration curve in the untransformed
scale is forced through the origin.
In addition, linear regression requires that the X-values are regularly
distributed over the range. This is usually not the case. After
logarithmic transformation this requirement may be fulfilled better (in
practice, again, the situation is probably somewhere between).
With respect to the use NONlinear regression and the monotonic model: Y
a*X**n + b: If data are not weighted, there is certainly a specified
underlying (hidden) assumption about the error model. I would not
recommend to use this method without careful analysis of the results,
e.g. in residuals plots, quite similar to common practice in
pharmacokinetics.
Best regards,
Hans
Johannes H. Proost
Dept. of Pharmacokinetics and Drug Delivery
University Centre for Pharmacy
Antonius Deusinglaan 1
9713 AV Groningen, The Netherlands
tel. 31-50 363 3292
fax 31-50 363 3247
Email: j.h.proost.-a-.farm.rug.nl
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Copyright 1995-2010 David W. A. Bourne (david@boomer.org)